為大家收集整理了《2014年計(jì)算機(jī)軟考系統(tǒng)分析師考前練習(xí)題》供大家參考，希望對(duì)大家有所幫助?。?！
●設(shè)∫x0f(t2)=2x3,則∫10f(x)=(1)。
（1）
A．1
B．2
C．3
D．4
●過原點(diǎn)做曲線y=ex的切線，則切線的方程為（2）。
（2）
A．y=ex
B．y=ex
C．y=x
D．y=ex/2
●lim(sin3x/tg2x)=(3)。
x→0
(3)
A．3
B．3/2
C．2
D．1
●命題公式￢（P∨Q）←→(P∧Q)的合取范式為（4），析取范式為（5）。
（4）
A．(P∨Q)∧(￢P∨￢Q)
B．(P∨￢Q)∧(￢P∨￢Q)
C．(￢P∨Q)∧(P∨￢Q)
D．(P∨Q)∧(￢P∨Q)
(5)
A．(P∧Q)∨(￢P∧Q)
B．(￢P∧￢Q)∨(￢P∧Q)
C．(P∧￢Q)∨(￢P∧Q)
D．(P∧￢Q)∨(P∧Q)
●In the following essay, each bland has four choices. Choose the best answer and write down on the answer sheet. With the implementation of (6) the so – called network has become a reality. The provision of such facilities is the most important part of the network requirements. However, in many applications the communicating computers may be of different types. This means that they may use different programming languages and, more importantly, different forms of (7) interface between user(application)program, normally referred to ,as application processes, and the underlying communication services may be (8). For example, one computer may be a small single-user computer, while another may be a large (9) system. In the earlier days of computer communication, these issues meant that only closed communities of computers (that is, from the same manufacturer) could communicate with each other in a meaningful way. IBM’s systems Network Architecture (SNA) and DEC’s Digital Network Architecture (DNA) are just two examples of communication software packages produced by manufacturers to allow their systems to be intercormected together. These proprietary packages, however, of which there am still many inexistence, do not address the problem of universal interconnect ability, or open systems interconnection. In an attempt to alleviate this problem,(10) ,in the late 1970s,formulated a reference model to provide a common basis (or the coordination of standards developments and to allow existing and evolving standards activities to be placed into perspective with one another.
(6)
A.communicationchannel
B.protocols
C.datachannel
D.publiccommunication
(7)
A.database
B.datarepresentation
C.protocols
D.datacommunication
(8)
A.same
B.similar
C.different
D.dependent
(9)
A.multi-user
B.client
C.server
D.full-user
(10)
A.DNA
B.SNA
C.theOSI
D.theISO
參考答案：
1、C 公式d(∫x0f(t))/dx=f(x),所以f(x)=6x,∫10f(x)=∫106xdx=3x2|x0=3
2、B 本題中f(x)=exf’(x)=ex設(shè)所求切線方程為y-ex0=ex0(x-x0).由于切線過原點(diǎn)，所以0-ex0=ex0(0-x0)，解得x0=1，故所求切線方程為y-e=e(x-1)即y=ex
3、B
lim(sin3x/tg2x)=lim(sin3x/sin2x)cos2x=lim((sin3x/3x)/(sin2x/2x))(3/2)cos2x=(1/1)*(3/2)*1=3/2
x→0x→0
4、A
￢（P∨Q）←→(P∧Q)
ó(￢（P∨Q）→(P∧Q))∧((P∧Q)→￢（P∨Q）)
ó(（P∨Q）∨(P∧Q))∧((￢P∨￢Q)∨(￢P∧￢Q))
ó(P∨Q)∧(￢P∨￢Q)
5、C
利用∧對(duì)∨分配得
(P∨Q)∧(￢P∨￢Q)
ó(P∧￢P)∨(P∧￢Q)∨（￢P∧Q）∨（Q∧￢Q）
ó(P∧￢Q)∨（￢P∧Q）
6、B 此處Protocol表示協(xié)議的意思，由協(xié)議的出現(xiàn)才使得網(wǎng)絡(luò)的運(yùn)行成為一種可能性，所以選擇B
7、B 在此表示數(shù)據(jù)的表示形式不同因此選B
8、C 這里與上文對(duì)應(yīng)，表示底層的不同通信設(shè)備由于協(xié)議的出現(xiàn)可以進(jìn)行相互的通信
9、A 與上文內(nèi)容對(duì)應(yīng)，上文提及single-user,是指單用戶，此處應(yīng)該是multi-user，多用戶的意思。
10、C OSI表示的是開放式系統(tǒng)互聯(lián)參考模型，符合文中含義，所以只能選C
●設(shè)∫x0f(t2)=2x3,則∫10f(x)=(1)。
（1）
A．1
B．2
C．3
D．4
●過原點(diǎn)做曲線y=ex的切線，則切線的方程為（2）。
（2）
A．y=ex
B．y=ex
C．y=x
D．y=ex/2
●lim(sin3x/tg2x)=(3)。
x→0
(3)
A．3
B．3/2
C．2
D．1
●命題公式￢（P∨Q）←→(P∧Q)的合取范式為（4），析取范式為（5）。
（4）
A．(P∨Q)∧(￢P∨￢Q)
B．(P∨￢Q)∧(￢P∨￢Q)
C．(￢P∨Q)∧(P∨￢Q)
D．(P∨Q)∧(￢P∨Q)
(5)
A．(P∧Q)∨(￢P∧Q)
B．(￢P∧￢Q)∨(￢P∧Q)
C．(P∧￢Q)∨(￢P∧Q)
D．(P∧￢Q)∨(P∧Q)
●In the following essay, each bland has four choices. Choose the best answer and write down on the answer sheet. With the implementation of (6) the so – called network has become a reality. The provision of such facilities is the most important part of the network requirements. However, in many applications the communicating computers may be of different types. This means that they may use different programming languages and, more importantly, different forms of (7) interface between user(application)program, normally referred to ,as application processes, and the underlying communication services may be (8). For example, one computer may be a small single-user computer, while another may be a large (9) system. In the earlier days of computer communication, these issues meant that only closed communities of computers (that is, from the same manufacturer) could communicate with each other in a meaningful way. IBM’s systems Network Architecture (SNA) and DEC’s Digital Network Architecture (DNA) are just two examples of communication software packages produced by manufacturers to allow their systems to be intercormected together. These proprietary packages, however, of which there am still many inexistence, do not address the problem of universal interconnect ability, or open systems interconnection. In an attempt to alleviate this problem,(10) ,in the late 1970s,formulated a reference model to provide a common basis (or the coordination of standards developments and to allow existing and evolving standards activities to be placed into perspective with one another.
(6)
A.communicationchannel
B.protocols
C.datachannel
D.publiccommunication
(7)
A.database
B.datarepresentation
C.protocols
D.datacommunication
(8)
A.same
B.similar
C.different
D.dependent
(9)
A.multi-user
B.client
C.server
D.full-user
(10)
A.DNA
B.SNA
C.theOSI
D.theISO
參考答案：
1、C 公式d(∫x0f(t))/dx=f(x),所以f(x)=6x,∫10f(x)=∫106xdx=3x2|x0=3
2、B 本題中f(x)=exf’(x)=ex設(shè)所求切線方程為y-ex0=ex0(x-x0).由于切線過原點(diǎn)，所以0-ex0=ex0(0-x0)，解得x0=1，故所求切線方程為y-e=e(x-1)即y=ex
3、B
lim(sin3x/tg2x)=lim(sin3x/sin2x)cos2x=lim((sin3x/3x)/(sin2x/2x))(3/2)cos2x=(1/1)*(3/2)*1=3/2
x→0x→0
4、A
￢（P∨Q）←→(P∧Q)
ó(￢（P∨Q）→(P∧Q))∧((P∧Q)→￢（P∨Q）)
ó(（P∨Q）∨(P∧Q))∧((￢P∨￢Q)∨(￢P∧￢Q))
ó(P∨Q)∧(￢P∨￢Q)
5、C
利用∧對(duì)∨分配得
(P∨Q)∧(￢P∨￢Q)
ó(P∧￢P)∨(P∧￢Q)∨（￢P∧Q）∨（Q∧￢Q）
ó(P∧￢Q)∨（￢P∧Q）
6、B 此處Protocol表示協(xié)議的意思，由協(xié)議的出現(xiàn)才使得網(wǎng)絡(luò)的運(yùn)行成為一種可能性，所以選擇B
7、B 在此表示數(shù)據(jù)的表示形式不同因此選B
8、C 這里與上文對(duì)應(yīng)，表示底層的不同通信設(shè)備由于協(xié)議的出現(xiàn)可以進(jìn)行相互的通信
9、A 與上文內(nèi)容對(duì)應(yīng)，上文提及single-user,是指單用戶，此處應(yīng)該是multi-user，多用戶的意思。
10、C OSI表示的是開放式系統(tǒng)互聯(lián)參考模型，符合文中含義，所以只能選C